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Exercise 4.3 · Q49

Q.△OAB\triangle OAB is formed by lines x2−4xy+y2=0x^2 - 4xy + y^2 = 0 and the line 2x+3y−1=02x + 3y - 1 = 0. Find the equation of the median of the triangle drawn from OO.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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From 2x+3y−1=02x+3y-1=0, y=1−2x3y=\tfrac{1-2x}3. Substituting into x2−4xy+y2=0x^2-4xy+y^2=0 and clearing denominators gives 37x2−16x+1=037x^2-16x+1=0, so x1+x2=1637x_1+x_2=\tfrac{16}{37}, and the midpoint's xx-coordinate is 837\tfrac{8}{37}. Since the midpoint lies on 2x+3y−1=02x+3y-1=0: 1637+3y=1⇒y=737\tfrac{16}{37}+3y=1 \Rightarrow y=\tfrac{7}{37}. So …

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